Number of stopping station problem (original) (raw)

Last Updated : 10 Nov, 2022

There are 12 intermediate stations between two places A and B. Find the number of ways in which a train can be made to stop at 4 of these intermediate stations so that no two stopping stations are consecutive?
Examples -

Input : n = 12, s = 4
Output : 126

Input : n = 16, s = 5
Output : 792

Explanation 1 :
Fix/remove of the four stops as fixed points and calculate in how many ways the other stations can be inserted between them, if you must have at least one station between stops.

        A   x   x   x   x   x   x   x   x   B

Between these 8 non-halting stations we have 9 places and we select these 9 places as halt between these 8 stations.
Therefore, answer should be ^{9}C_4 = 126

Explanation 2 :
If you know about combinations about indistinguishable balls into distinguishable boxes, then you can simply use, ^{(n-p+1)}C_p . In this question, nnn is number of stations and ppp is number of stations on which you want to stop. Here stopping stations are as indistinguishable balls and non-stopping stations are as distinguishable boxes.
Therefore, ^{(12-8+1)}C_4 = ^{9}C_4 = 126

Below is the code implementation of the solution:

C++ `

#include <stdio.h> int stopping_station(int, int);

// function to calculate number // of ways of selecting 'p' non consecutive // stations out of 'n' stations int stopping_station(int p, int n) {

int num = 1, dem = 1, s = p;

// selecting 's' positions out of 'n-s+1'
while (p != 1) {

    dem *= p;
    p--;
}

int t = n - s + 1;
while (t != (n - 2 * s + 1)) {

    num *= t;
    t--;
}

if ((n - s + 1) >= s)
    printf("%d", num / dem);

else

    // if conditions does not satisfy of combinatorics
    printf("not possible");

}

// driver code int main() {

// n is total number of stations
// s is no. of stopping stations
int n, s;

// arguments of function are
// number of stopping station
// and total number of stations
stopping_station(4, 12);

}

Java

// Java code to calculate number // of ways of selecting 'p' non // consecutive stations out of // 'n' stations

import java.io.; import java.util.;

class GFG { public static int stopping_station(int p, int n) { int num = 1, dem = 1, s = p;

    // selecting 's' positions out of 'n-s+1'
    while (p != 1)
    {
        dem *= p;
        p--;
    }

    int t = n - s + 1;
    while (t != (n - 2 * s + 1))
    {
        num *= t;
        t--;
    }

    if ((n - s + 1) >= s)
        System.out.print(num / dem);

    else
        // if conditions does not satisfy of combinatorics
        System.out.print("not possible");

    return 0;
}

public static void main (String[] args)
{
    // n is total number of stations
    // s is no. of stopping stations
    int n, s;

    // arguments of function are
    // number of stopping station
    // and total number of stations
    stopping_station(4, 12);
}

} // ""This code is contributed by Mohit Gupta_OMG ""

Python3

Python code to calculate number

of ways of selecting 'p' non

consecutive stations out of

'n' stations

def stopping_station( p, n): num = 1 dem = 1 s = p

# selecting 's' positions
# out of 'n-s+1'
while p != 1:
    dem *= p
    p-=1

t = n - s + 1
while t != (n-2 * s + 1):
    num *= t
    t-=1
if (n - s + 1) >= s:
    return int(num/dem)
else:
    # if conditions does not
    # satisfy of combinatorics
    return -1

driver code

num = stopping_station(4, 12) if num != -1: print(num) else: print("Not Possible")

This code is contributed by "Abhishek Sharma 44"

C#

// C# code to calculate number // of ways of selecting 'p' non // consecutive stations out of // 'n' stations using System;

class GFG {

public static int stopping_station(int p, 
                                   int n)
{
    int num = 1, dem = 1, s = p;

    // selecting 's' positions 
    // out of 'n-s+1'
    while (p != 1)
    {
        dem *= p;
        p--;
    }

    int t = n - s + 1;
    while (t != (n - 2 * s + 1))
    {
        num *= t;
        t--;
    }

    if ((n - s + 1) >= s)
        Console.WriteLine(num / dem);

    // if conditions does not 
    // satisfy of combinatorics
    else
      Console.WriteLine("Not possible");
    return 0;
}

// Driver Code
public static void Main(String []args)
{
    
    // arguments of function are
    // number of stopping station
    // and total number of stations
    stopping_station(4, 12);
}

}

// This code is contributed by vt_m.

PHP

JavaScript

`

Output :

126

Time Complexity: O(s)
Auxiliary Space: O(1)